Difference between revisions of "2013 AIME II Problems/Problem 11"

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==Problem 11==
 
==Problem 11==
 
Let <math>A = \{1, 2, 3, 4, 5, 6, 7\}</math>, and let <math>N</math> be the number of functions <math>f</math> from set <math>A</math> to set <math>A</math> such that <math>f(f(x))</math> is a constant function. Find the remainder when <math>N</math> is divided by <math>1000</math>.
 
Let <math>A = \{1, 2, 3, 4, 5, 6, 7\}</math>, and let <math>N</math> be the number of functions <math>f</math> from set <math>A</math> to set <math>A</math> such that <math>f(f(x))</math> is a constant function. Find the remainder when <math>N</math> is divided by <math>1000</math>.
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==Solution==
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Any such function can be constructed by distributing the elements of <math>A</math> on three tiers.
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The bottom tier contains the constant value, <math>c=f(f(x))</math> for any <math>x</math>. (Obviously <math>f(c)=c</math>.)
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The middle tier contains <math>k</math> elements <math>x\ne c</math> such that <math>f(x)=c</math>, where <math>1\le k\le 6</math>.
  
==Solution==
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The top tier contains <math>6-k</math> elements such that <math>f(x)</math> equals an element on the middle tier.
{{solution}}
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There are <math>7</math> choices for <math>c</math>. Then for a given <math>k</math>, there are <math>\tbinom6k</math> ways to choose the elements on the middle tier, and then <math>k^{6-k}</math> ways to draw arrows down from elements on the top tier to elements on the middle tier.
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Thus <math>N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399</math>, giving the answer <math>\boxed{399}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=10|num-a=12}}
 
{{AIME box|year=2013|n=II|num-b=10|num-a=12}}
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{{MAA Notice}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 19:49, 18 August 2021

Problem 11

Let $A = \{1, 2, 3, 4, 5, 6, 7\}$, and let $N$ be the number of functions $f$ from set $A$ to set $A$ such that $f(f(x))$ is a constant function. Find the remainder when $N$ is divided by $1000$.

Solution

Any such function can be constructed by distributing the elements of $A$ on three tiers.

The bottom tier contains the constant value, $c=f(f(x))$ for any $x$. (Obviously $f(c)=c$.)

The middle tier contains $k$ elements $x\ne c$ such that $f(x)=c$, where $1\le k\le 6$.

The top tier contains $6-k$ elements such that $f(x)$ equals an element on the middle tier.

There are $7$ choices for $c$. Then for a given $k$, there are $\tbinom6k$ ways to choose the elements on the middle tier, and then $k^{6-k}$ ways to draw arrows down from elements on the top tier to elements on the middle tier.

Thus $N=7\cdot\sum_{k=1}^6\tbinom6k\cdot k^{6-k}=7399$, giving the answer $\boxed{399}$.

See Also

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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